如何重复使用block_diag [英] How to use block_diag repeatedly

问题描述

我有一个非常简单的问题，但仍然无法解决. 我想要一个块对角n ^ 2 * n ^ 2矩阵.块是只有对角线的稀疏n * n矩阵，首先是对角线，然后是对角线.对于n=4的简单情况，可以轻松完成

I have rather simple question but still couldn´t make it work. I want a block diagonal n^2*n^2 matrix. The blocks are sparse n*n matrices with just the diagonal, first off diagonals and forth off diag. For the simple case of n=4 this can easily be done

datanew = ones((5,n1))
datanew[2] = -2*datanew[2]
diagsn = [-4,-1,0,1,4]
DD2 = sparse.spdiags(datanew,diagsn,n,n)
new = sparse.block_diag([DD2,DD2,DD2,DD2])

由于这仅对小n有用，是否有更好的方法来使用block_diag? n-> 1000的思考

Since this only useful for small n's, is there a way better way to use block_diag? Thinking of n -> 1000

推荐答案

构造DD2矩阵长列表的一种简单方法是使用列表理解:

A simple way of constructing a long list of DD2 matrices, is with a list comprehension:

In [128]: sparse.block_diag([DD2 for _ in range(20)]).A
Out[128]: 
array([[-2,  1,  0, ...,  0,  0,  0],
       [ 1, -2,  1, ...,  0,  0,  0],
       [ 0,  1, -2, ...,  0,  0,  0],
       ..., 
       [ 0,  0,  0, ..., -2,  1,  0],
       [ 0,  0,  0, ...,  1, -2,  1],
       [ 0,  0,  0, ...,  0,  1, -2]])

In [129]: _.shape
Out[129]: (80, 80)

至少在我的版本中，block_diag想要一个数组列表，而不是*args:

At least in my version, block_diag wants a list of arrays, not *args:

In [133]: sparse.block_diag(DD2,DD2,DD2,DD2)
...
TypeError: block_diag() takes at most 3 arguments (4 given)

In [134]: sparse.block_diag([DD2,DD2,DD2,DD2])
Out[134]: 
<16x16 sparse matrix of type '<type 'numpy.int32'>'
    with 40 stored elements in COOrdinate format>

这可能不是构造这样的块对角线数组的最快方法，但这是一个开始.

This probably isn't the fastest way to construct such a block diagonal array, but it's a start.

================

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看看sparse.block_mat的代码，我推断它确实可以做到:

Looking at the code for sparse.block_mat I deduce that it does:

In [145]: rows=[]
In [146]: for i in range(4):
    arow=[None]*4
    arow[i]=DD2
    rows.append(arow)
   .....:     

In [147]: rows
Out[147]: 
[[<4x4 sparse matrix of type '<type 'numpy.int32'>'
    with 10 stored elements (5 diagonals) in DIAgonal format>,
  None,
  None,
  None],
 [None,
  <4x4 sparse matrix of type '<type 'numpy.int32'>'
  ...
  None,
  <4x4 sparse matrix of type '<type 'numpy.int32'>'
    with 10 stored elements (5 diagonals) in DIAgonal format>]]

换句话说，rows是None的矩阵"，沿着对角线是DD2.然后将它们传递给sparse.bmat.

In other words, rows is a 'matrix' of None with DD2 along the diagonals. It then passes these to sparse.bmat.

In [148]: sparse.bmat(rows)
Out[148]: 
<16x16 sparse matrix of type '<type 'numpy.int32'>'
    with 40 stored elements in COOrdinate format>

bmat依次从所有输入矩阵的coo格式中收集data,rows,cols，将它们连接到主数组中，并从它们中构建一个新的coo矩阵.

bmat in turn collects the data,rows,cols from the coo format of all the input matricies, joins them into master arrays, and builds a new coo matrix from them.

因此，一种替代方法是直接构造这3个数组.

So an alternative is to construct those 3 arrays directly.

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